Question

A plane $P=0$ is the perpendicular bisector of the line joining the points $\left(2,3 , 4\right)$ and $\left(6,7 , 8\right).$ The perpendicular distance of $P=0$ from the origin is

• A. $4\sqrt{3}$ units
• B. $5\sqrt{3}$ units
• C. $6\sqrt{3}$ units
• D. $8\sqrt{3}$ units

Answer 1

Answer: (B)

From diagram mid-point of $AB$ must lie on the plane
$M\left(\frac{2 + 6}{2} , \frac{3 + 7}{2} , \frac{4 + 8}{2}\right)\equiv \left(4,5 , 6\right)$
$AB$ is perpendicular to the required plane
Direction ratios of $\overset{ \rightarrow }{A B}= < 4,4,4>= < 1,1,1>$
Equation of the required plane is
$1\left(x - 4\right)+1\left(y - 5\right)+1\left(z - 6\right)=0$
$\Rightarrow x+y+z=15$
The perpendicular distance from the origin to the plane $=\left|\frac{0 + 0 + 0 - 15}{\sqrt{3}}\right|=5\sqrt{3}$ units